taylor expansion of an integral $\int_0^1{e^{x^2}}$

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You should use a different form of the remainder term, namely the tail of the series: $$R_n(x)=\sum_{k=n+1}^\infty \frac{x^{2k}}{k!}$$ and then you estimate that, for example by comparing with a suitable geometric series, for example $$\frac{x^{2n+2}}{(n+1)!}\sum_{j=0}^\infty\Bigl(\frac{x^2}{n+2}\Bigr)^{j}.$$

By the way, you can also integrate the series for $e^{x^2}$ and get a new series for $\int_0^x e^{t^2}\,dt$ and use that directly.

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Thomas Russell
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Thomas Russell

Updated on May 02, 2020

Comments

  • Thomas Russell
    Thomas Russell over 3 years

    I need to calculate $\int_0^1{e^{x^2}\:dx}$ with taylor expasin in accurancy of less than 0.001. The taylor expansion around $x_0=0$ is $e^{x^2}=1+x^2+\frac{x^4}{3!}+...$. I need to calculate when the rent is smaller than 0.001. So, I got $R_n(x)=\frac{f^{(n+1)}(c) x^{n+1}}{(n+1)!}$. the function and its derivaties are monotonically increasing so $| \int R_n(x)dx|\leq \int|R_n(x)dx|\leq \int \frac{f^{(n+1)}(1) x^{n+1}}{(n+1)!} \leq10^{-3}$. I need to find the value of $f^{(n+1)}(1)$. How can I do that?

  • Admin
    Admin over 10 years
    Doesn't the tail of series includes also exprsssion from the derivative? about geometric series do you mean $sum_{k=n+1}^{\infty}x^{2k}$?
  • Harald Hanche-Olsen
    Harald Hanche-Olsen over 10 years
    There are several points here: Directly computing higher derivatives of $e^{x^2}$ is prohibitively difficult. It is of course true that you can compute higher derivatives from the series, and all information about the $n$th derivative is in the $n$th tail of the series. But that would be a lot of unnecessary work, given that it is the remainder you want in the end; the derivative would only be a means to that end, and one that is actually not needed. Regarding what geometric series, I'll add that to the answer.
  • Admin
    Admin over 10 years
    about main content:Last time I did it in exam the lecturer gave me 30% of the mark for the question about taylor series (great times of calculus 1...) about the aside:you are right. So which upper bound can I find for $f^{(n+1)}(c)?$